Hankel forms and Nehari's theorem

Transcription

1 Hankel forms and Nehari's theorem Øistein Søvik Master of Science Submission date: May 27 Supervisor: Kristian Seip, IMF Co-supervisor: Ole Fredrik Brevig, MAH Norwegian University of Science and echnology Department of Mathematical Sciences

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3 Abstract. he purpose of this thesis is to explore the relation between the classical Hardy space of analytic functions and the Hardy space of Dirichlet series. wo chapters are devoted to developing the basic properties of these spaces. In the remaining two chapters we study Nehari s theorem both in the classical and multiplicative setting as a concrete example of the usefulness of the interplay between the space of Dirichlet series and the space of analytic functions on the infinite-dimensional polydisc. Sammendrag. Formålet med denne oppgaven er å utforske sammenhengen mellom Hardy rommet av analytiske funksjonene i polydisken og Hardy rommet av Dirichlet rekker. o kapittler er satt av til å utforske egenskapene til disse rommene. I de to gjenværende kapittlene studeres Neharis teorem både i den klassiske og multiplikative settingen som et konkret eksempel på nytterdien av å utnytte samspillet mellom rommet av Dirichlet rekker og rommet av analytiske funksjoner på den uendelig-dimensjonale polydisken. iii

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5 Preface his thesis was written from January to June 27 under supervision of Ole Fredrik Brevig, and marks the end of my time as a student at the Department of Mathematical Science and at the eacher Education program at NNU. I thank Brevig for suggesting the fascinating topic at hand, and have found working with this thesis to be very educative, and it has allowed me to focus on the parts of mathematics that I have come to enjoy the most. I am also indebted to Brevig for taking his time to meet me twice a week, providing detailed feedback feedback on my drafts, and offering enlightening discussions when things looked dark. In addition, a big thanks to my friends here in rondheim, who have made these years very memorable, and the studies that much easier. Finally, thank you to my family for always supporting me and building me up. While there is nothing groundbreaking in this thesis, I feel like I have given the topic a coherent treatise from the classical to the cutting-edge results, while simplifying some proofs in the process. All in all I am fairly satisfied with the end product. I hope you enjoy your reading. Øistein Søvik rondheim, 27 v

6 Contents Abstract Sammendrag Preface Contents Symbols and abbreviations iii iii v viii ix Introduction Hankel forms and Dirichlet series Overview of the thesis 3 Chapter. Hardy spaces on the disc 5.. Preliminaries 5.2. he Hardy space 9.3. he zeroes of functions in H p 2.4. Boundary functions 9.5. Carleman s inequality 2.6. Hardy spaces on the polydisc Helson s inequality 27 Chapter 2. Hankel forms Bilinear forms he Hilbert matrix Nehari s theorem and weak product spaces 36 Chapter 3. he Hardy space of Dirichlet series Preliminaries he Hardy-Hilbert space H he Hardy space H p 5 Chapter 4. Multiplicative Hankel forms he multiplicative Hilbert matrix Nehari s theorem Hilbert-Schmidt forms Some related open problems 68 vi

7 CONENS vii Appendix A. Preliminaries 7.. Euler products 7.2. Inequalites Integrals and sums Functional analysis and measure theory 85 Bibliography 87

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9 Symbols and abbreviations Symbol Description Page N he natural numbers:, 2, 3, Z he integers:..., 3, 2,,,, 2, 3, R he real numbers. 5 D he unit disk. 5 Boundary of the unit disk. 5 o(f(n)) Little O-notation: If g o(f(x)) then g/f as x. 35 O(f(n)) Big O-notation: g O(f(x)) if and only if there exists a positive 34 real number C and a real number k such that g(x) Cf(x) for all x k. dm Normalized Lebesgue measure on such that m() =. 5 dσ Normalized Lebesgue measure on D such that σ(d) =. 2 L p he space of Lebesgue integrable functions. 5 A p he Bergman space of analytic functions. 2 B(z) he Blaschke product. 4 C he complex plane {σ + it : σ, t R}. 8 C θ he complex half plane {σ + it: σ > θ, t R}. 44 ix

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11 Introduction We begin by a short introduction to the topic at hand, through Hankel forms and Dirichlet series. After this a short overview of each chapter is given. Hankel forms and Dirichlet series A Hankel form in l 2 l 2 C is one of the form ρ(a, b) := a m b m ρ m+n, m,n and we say that the Hankel form is bounded if there exists a constant such that ( ) ( 2 ) 2 a m b n ρ n+m C a m 2 b n 2. m,n= m= Further we let H 2 (D) denote the Hilbert space of functions analytic in D with square-summable aylor coefficients. Every function ψ = j ρ jz j in H 2 (D) defines a Hankel form H ψ by the relation n= H ψ (fg) = fg, ϕ H 2, f, g H 2. he most important theorem for Hankel forms is the Nehari s theorem [33], which states that every bounded Hankel form is generated by a bounded symbol ψ on the torus. More precisely H ψ extends to a bounded form on H 2 () H 2 () if and only if ψ = P + ϕ for a bounded function ϕ in L (). Where P + denotes the orthogonal projection L 2 () H 2 (). An interesting question is whether the multiplicative Hankel forms ϱ(a, b) := a m b m ϱ mn, m,n exhibits the same properties as the (additive) Hankel forms. hese can be viewed as the classical Hankel forms now on the infinite-dimensional polydisc. We let H 2 denote the Hilbert space of Dirichlet series with square-summable coefficients in the half plane C /2 = { s C, Re s > /2 }. Every Dirichlet series ψ = n ρ nn s in H 2 defines a multiplicative Hankel form H ψ by the relation H ψ (f, g) = fg, ψ H 2, f, g H 2, he main purpose of this thesis is to explore to what extent Nehari s theorem holds for these multiplicative Hankel forms. his study is started by exploring the properties of the multiplicative analog of the Hilbert matrix whose analytic symbol

12 2 INRODUCION ϕ is the primitive of ζ(s + /2) in H 2. As shown in [] this Hankel form is bounded with norm π. More explicitly written n,m 2 a m b n ( π nm log(nm) n 2 a m 2) ( 2 b n 2) 2 his raises the following question. Question. Does the multiplicative Hilbert matrix have a bounded symbol? o which the answer is still maybe. A key tool in the study of Dirichlet series and Hardy spaces is the Bohr lift. For any n N, the fundamental theorem of arithmetic yields n = j p κj j, which associates the finite non-negative multi-index κ(n) = (κ, κ 2, κ 3,...) to n. he Bohr lift of the Dirichlet series is the power series Bf(z) = n a n z κ(n), where z = (z, z 2, z 3,...). Under the Bohr lift, a formal computation shows that n 2 BfBg, Bϕ L 2 ( ) = fg, ϕ H 2, allowing us to compute the multiplicative Hankel form on the infinite polydisk. he study of Hankel forms on was initiated by Helson [25, p ], who raised the following questions: Question 2. Does every bounded multiplicative Hankel form have bounded symbol ϕ on the polytorus? Question 3. Does every multiplicative Hankel form in the Hilbert Schmidt class have a bounded symbol? We answer these questions in full detail chapter 4. By realizing Hankel forms as small operators on the polydisk and using ideas from Ortega-Cerdà and Seip [36], Bayart et al. [6], and Brevig and Perfekt [9] we answer Question in the negative. By extending Carleman s inequality into the polydisk we obtain Helson s inequality, and using this inequality we prove that every multiplicative Hankel form in the Hilbert Schmidt class have a bounded symbol, thus confirming Question 3.

13 OVERVIEW OF HE HESIS 3 Overview of the thesis Chapter. he first chapter is an introduction to the classical Hardy spaces on the disk. We prove the Riesz factorization theorem, and use the results to show that the space of polynomials are dense in H p. We then extend the properties of the point estimate and Carleman s inequality to the polydisk; these results are respectively known as the Cole-Gamelin estimate and Helson s inequality. Chapter 2. he second chapter introduces the Hankel forms and shows their relationship with functions in the Hardy space H 2. We study the bona fide example of a Hankel form, namely the Hilbert matrix. hen we use the weak-factorization of the Hardy space on the disk to prove Nehari s theorem for Hankel forms. Chapter 3. In the third chapter, we study the Hardy space of Dirichlet series, and prove that this space behaves similarly to the Hardy spaces. In particular we prove Carlson s theorem, and use it to show that H 2 is the closure of Dirichlet polynomials under the Besicovitch norm. Using an idea of Brevig and a bilinear form, a sharp estimate for an embedding inequality is obtained. he Bohr correspondence is then introduced, and we use it to obtain the pointestimate for H p. Using the Bohr correspondence and idea of Saksman and Seip we offer an elementary proof that Hardy space H p may be defined as the Banach space completion of Dirichlet polynomials in the Besicovitch norm, thus extending Carleson s theorem. Chapter 4. In the last chapter we introduce the multiplicative Hankel forms, and study the multiplicative analogue to the Hilbert matrix. We prove that this Hankel form is bounded with same norm as the Hilbert matrix. he chapter ends by proving that Nehari s theorem does not hold in full generality, this is done by studying Hankel forms as small operators on the polydisc. We also show that Nehari s theorem holds under the restriction that the symbol is completely multiplicative or has square summable coefficients.

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15 CHAPER Hardy spaces on the disc his chapter begins with some preliminaries, before the classical definition of the Hardy space is presented together with some basic results on boundary behavior. his work is done in preparation for proving the Riesz factorization theorem, which has a number of interesting applications. In particular we show that every function f H can be written as f = gh, where g, h H 2 and f = g 2 h 2. At the end we show that the Hardy spaces may be defined as the closure of the polynomials in L 2, and extend some of our results to the polydisc... Preliminaries his section is devoted to introducing a series of necessary prerequisites. In particular we briefly introduce harmonic functions, the Poisson kernel and Möbius transformations. In particular we need some results about the radial limits of harmonic, and thus also analytic functions. For brevity the proofs are omitted, see Pavlović [38], Rudin [44, Chp. XI], or Duren [5, Chp. I] for reference. Following the notation of standard literature we will denote the unit disk D as D = { z C: z < }. Similarly, rather than D will represent the boundary of the disk = { z C: z = } = {e it : t R/2πZ}. Functions defined on will be identified with functions on R/2πZ, i.e. with functions on the real line, periodic of period 2π. Here Z denotes the set of integers {...,,,,...}, and similarly N represents the set positive integers {, 2,...}. Integrals on will be with respect dm= dθ/2π, the normalized Lebesgue measure such that m() =. We will use the following notations to describe integrals on and over the real numbers R: f dm := 2π π π f(e iθ ) dθ and R f dx := f(x) dx, and the notation f(z) dm(z) will be used whenever the need to specify which variable we are integrating over arises. Similarly, the notation f(n) = f(n) and f(n, m) = f(n, m), n Z n= n= m= m,n will frequently be used, and the latter expression will naturally be extended to as many variables as needed. For p <, we will let L p () denote the Banach 5

16 6. HARDY SPACES ON HE DISC space consisting of all analytic functions satisfying ( f Lp () := f ) p p dm <. When p =, we define L as the space of essentially bounded functions f L () := f(e iθ ). sup θ<2π For brevity we will write L p = L p () when no confusion is possible.... Harmonic functions and the Poisson kernel Definition (Harmonic functions). Let u be an analytic function in an open set Ω, such that u 2 / 2 x and u 2 / 2 y exists at every point of Ω. he Laplacian of u is defined as u := 2 u x u y 2. If u C 2 (Ω) is a twice continuously differentiable function in Ω and if u =,, at every point of Ω, then u is said to be harmonic in Ω. heorem.. A harmonic function u defined on a simply connected domain Ω can be represented in the form u(z) = h(z)+g(z), z Ω, where h and g are analytic and uniquely determined up to an additive constant; conversely, if u = h + g, where h and g are analytic, then f is harmonic. Using this theorem one can deduce various properties of harmonic functions from the corresponding properties of analytic functions and vice versa. he Poisson integral and kernel. One of the most used and well known harmonic functions is the Poisson kernel, see [, p ], [44, p. -2, Chp. XI] or Pavlović [38, Chp. III] for futher details. Definition. For all r < and θ [, 2π), the Poisson kernel is defined as P r (θ) := r n e inθ r 2 = 2r cos θ + r 2. (.) n= Definition. he Poisson integral of a function φ L p (φ) is the harmonic function P [φ] defined by P [φ] := P r φ := P r (t θ)ϕ(e iθ ) dm (re iθ D). (.2) he notation f g is referred to as the convolution of f and g. Perhaps the most useful property of the Poisson integral is that it can be used to solve the Dirichlet problem for the disk: heorem.2. If ϕ is a continuous function defined on, then ϕ has a unique continuous extension to D that is harmonic in D; this extension equals P [φ]. An immediate consequence is that set of all trigonometric polynomials is dense in each of the spaces C(), L p (), this known Weierstrass approximation theorem.

17 .. PRELIMINARIES he harmonic Hardy spaces Definition. Let p. We denote by h p the space of harmonic functions in D such that h p := { f : u h p < }. (.3) Here u h p is the norm of u, and defined as ( u h p := sup r< 2π f r p dm) /p, where the shorthand notation f r (e iθ ) = f(re iθ ) was introduced. In the case p = the integral is to be interpreted as a supremum: u h := sup u(z). z D hat heorem.2 extends to < p is shown in the following theorem: heorem.3. he function u belongs to h p ( < p ) if and only if it is equal to the Poisson integral of some function φ L p. And if f = P [φ], then u h p = φ L p. heorem.4 (Fatou s heorem [6], 96). Let u h p ( < p ), then u has a radial limit at almost every point e iθ. In particular lim f(reiθ ) = φ(e iθ ) for almost every θ [, 2π). r For a modern proof see Nikolski [34, p. 39]. he case L is treated in Rudin [44, p. 244], and Duren [5, p. 5]. While heorem.3 fails to hold for p =, the following is true: Corollary.5. Each function u h has a radial limit almost everywhere. Corollary.6. If u is the Poisson integral of a function ϕ L, then u(re iθ ) ϕ(θ) almost everywhere. As an example let φ = m= a me imθ be a function such that φ L. hen ( P [φ](re iθ ) = = n= m,n= r n e I(θ t)n)( a m e int m= e I(m n)θ dm = a m e imt) dt n= r n a n e int, So the operator P : L h is injective, as every function φ L may be seen as a boundary function of a function u h. However P is not onto as there exists functions in h, whose boundary function does not lie in L.

18 8. HARDY SPACES ON HE DISC..3. Subharmonic functions As usual a domain is an open connected set in the complex plane. Definition. A real-valued function g(z) is said to be subharmonic if it has the following property: For each domain B with B D, and for each function U(z) harmonic in B, continuous in the closure B, such that g(z) U(z) for B, then holds throughout B. g(z) U(z) In particular if there is a function U(z) harmonic in B with boundary values g(z), then g(z) U(z) in B. Proposition.7. If f is analytic in a domain D and p >, then f p is subharmonic in D. A transformation of the form..4. he Möbius group (z) = az + b cz + d, (.4) where a, b, c, d C and ab cd, is called a Möbius transformation. Where C = {σ + it: σ, t R} denotes the complex plane. Proposition.8. he Möbius transformation is a conformal one-to-one mapping that sends circles and lines to circles or lines. Before moving on we would like to present two useful Möbius transformations. he shifted Cayley transformation (z) = a + + z z, is a conformal one-to-one mapping of the open unit disk onto the open half plane C a. In particular if z lies on the boundary we have For any α D, define (e it ) = a + i tan(t/2). ϕ α (z) = z α αz. (.5) Fix α D. hen ϕ α is a one-to-one mapping which carries onto, D onto D and α to. We have φ a(z) = a 2 ( az) 2. (.6) Proposition.9 ( [44, hm. 2.6] ). Suppose is an Möbius transformation (ϕ is one-to-one, ϕ(d) = D, α D, and ϕ(α) = ). hen there exists a constant θ [, 2π), such that (z) = e iθ ϕ α (z) z D. (.7) In other words, we obtain (z) by composing the mapping ϕ α with a rotation. his mapping is also referred to as a linear fractional transformation.

19 .2. HE HARDY SPACE 9.2. he Hardy space In 95 Godfrey Harold Hardy, published in the Proceedings of the London Mathematical Society a paper confirming a question posed by Landau [9]. In this paper, not only did Hardy generalize Hadamard s three-circle theorem, but he also put in place the first brick of a new branch of mathematics which bears his name: the theory of Hardy spaces H p. For three decades afterwards mathematicians such as Hardy, Littlewood, Pólya, Riesz, Privalov, F. and V. Smirnov, and G. Szegö, expanded and developed the theory of the Hardy spaces. While most of this early work is concerned with properties of individual functions of class H p, the development of functional analysis has stimulated a new interest in the H p classes. For the interested reader an excellent exposition of the classical Hardy space is the monograph by Duren [5], other sources includes [29, 34] and the short treatise by Rudin [44, Chp. XVII]. In this section we shall look at properties of spaces which are represented by power series in D, i.e functions of the form f(z) = n a n z n, z = re iθ. (.8) When the power series in equation (.8) converges we call f an analytic function. As before we will work in the unit disk r <, and similar to how the f L p norm was defined, we introduce f H p := ( sup r< fr p dm ) /p = sup r< f r L p. (.9) and when p =, we use let the norm be defined as the essential supremum of f: f H := sup f(z). (.) z D Definition. Let p, the Hardy space H p (D) consists of those analytic functions in the unit disk D such that, f H p < +. As we will only work on the unit disk D will omit the domain and simply write H p when no confusion is possible. We will first look at the particular case p = 2 and then extend the properties to p..2.. he Hardy space H 2. With the norm of H 2 defined as above, the definition of the inner-product follows naturally: f, g 2 H := lim f 2 r g r dm = lim f r, g r 2 L. r 2 r In addition, we introduce the notation f (e iθ ) := lim r f(re iθ ). essential properties of H 2 are encapsulated in the following theorem: he most heorem.. Let f = n a nz n and g = n b nz n be analytic for z <, where z = re iθ. hen () f, g 2 H 2 = n a nb n.

20 . HARDY SPACES ON HE DISC (2) f H 2 = n a n 2 = f L 2. (3) f r L 2 is a non-decreasing function of r. (4) H 2 is a Hilbert space. (5) f(z) f H 2/ z 2. It will be convenient to first prove the following lemma. Lemma.. Let z, then {z j } j forms an orthonormal set in L 2. Proof. We start by introducing the Kronecker delta symbol δ jk, defined as if j = k, and otherwise. Proving the lemma is the same as showing z j, z k L 2 = z j z k dm = δ jk, for every j, k N. Since z we can write z = e iθ, and our integral becomes z j z k dm = 2π e iθ(j k) dθ. 2π It is clear that the integral is whenever j = k. Assume therefore that j k, 2π 2π e iθ(j k) dθ = e 2πi(j k) 2πi j k which completes the proof since e 2πi(j k) = for every integer pair j k. Proof of heorem.. We begin by applying Lemma. to the inner product of f = n a nz n and g = n b nz n : f r, g r 2 L = f 2 r g r dm = a n b m r n+m z n z m dz = a n b n r 2n, n,m n his proves that the inner product is increasing as a function of r, thus proving 3. Since < r <, we can apply the monotone convergence theorem on f r, g r 2 L to 2 obtain item. he computation above also shows f r, f r L 2 = f r 2 L 2 = m a m 2 r 2n, (.) and proves the first part of 2. Since L 2 () is a complete space, f L 2 () and we can compute the Fourier coefficients to be 2π { f (n) = f (e iθ nθ dθ 2π )e 2π = lim f r (e iθ inθ dθ )e r 2π = a n : n : n <, he second equality follows from the monotone convergence theorem since f is increasing. Combining this with Parseval s theorem shows f L 2 = a n 2 = f H 2, n

21 .2. HE HARDY SPACE thus completing the proof of 2. o prove 4 we need to show that every sequence f r f, as r is Cauchy in H 2 (). 2 Using Lemma. from above, and obvious modifications, f r f s 2 L = ( a 2 n r n s n) z n 2 dm = r n n ( n s n) a n 2. But as n a n 2 < we get by the dominated convergence theorem that the last summand goes to zero when r, s. hus, H 2 is a complete Hilbert space as f (e iθ ) = f(e iθ ) almost everywhere. o prove that point-wise evaluation of functions in H 2 is a bounded functional we may apply the Cauchy Schwarz inequality f(z) n a n z n ( n z 2n ) ( 2 a n 2 n ) 2 z 2 f H 2. where the last equality followed from applying item 2 and the geometric series n rn = /( r). his proves 5, and completes the proof of heorem.. From the preceding discussion we see that the polynomials are dense in H 2, thus the mapping f f establishes an isometry between H 2 and the closure of the polynomials in L 2. Hence, H 2 may be defined as as: () the set of analytic functions f in D such that lim r f r 2 dm <. (2) the closure of the polynomials in L 2 (). hat H p can be seen as the closure of the polynomials in L p and that heorem. can be extended to p is true, but not entirely trivial. A key part in proving this will the the Riesz factorization theorem. A stepping stone in proving this is the following theorem. he class H p was introduced as the set of all functions f(z) analytic in z < for which the means f r L p are bounded. As seen from., f r L 2 is increasing as a function of r, and the case p = is trivial as f r L increases with r from the maximum modulus principle. A natural question is therefore whether f r L p is always a non-decreasing function of r. his was proven by Hardy [9] and is considered the starting point of the theory of Hardy spaces. heorem.2 (Hardy s convexity theorem). For z < let f(z) be analytic, and let p. hen f r L p is a non-decreasing function of r. Proof. As pointed out in section..3 f p ( p ) is subharmonic if f is analytic. So it is enough to prove heorem.2 for subharmonic functions. Let g(z) be subharmonic in z <, and define m(r) := g r dm, r <. Choose r < r 2 <. Since g(z) is subharmonic there exists a function U such that, U(z) is harmonic in z < r 2, continuous in z r 2, and equal to g(z) for 2 hat we may associate H 2 () with a subspace of L 2 () follows from 2, and Fatou s heorem.4

22 2. HARDY SPACES ON HE DISC z = r 2. Hence, g(z) U(z) for z r 2, so m(r ) U r dm = U() = U r2 dm = m(r 2 ), by the mean-value property A.23. his proves that m(r) is non-decreasing, and so u L p is also non-decreasing. While not needed, it is also true that log f r L p is a convex function of log r, see Hardy [9] or Duren [5, p. 9]. Remark. heorem.2 implies we may replace the sup in the definition of the H p with a limit f H p = ( ) lim f r p p dm, r as the norm is increasing. he proof for p = follows again from the maximum modulus principle..3. he zeroes of functions in H p Let f L p, ( p ). We denote the zero sequence of f as Z (f) consisting of the elements { z D: f(z) = }, (.2) in increasing order of magnitude. It is well known that for a analytic function in the unit disk, either Z (f) = D or Z (f) has no limit points in D. he first case bears little interest as by the maximum modulus principle it implies f. hus, the zeroes of a non-zero analytic function f L p are isolated points in, and if the number of zeroes is infinite, the limit points have to lie outside D i.e. on the boundary. From the theorem of Weierstrass [44, Chapter 5] this is all we can say about the zeroes of analytic functions. However, if we instead consider functions in H p we can say much more about the distribution of zeroes in D, namely that the zeroes have to converge with a certain rate toward the limit points on. he basis of deriving the rate of conversion of the zeroes of H p is the following formula. heorem.3 (Jensen s Formula). Let f be an analytic function in a region which contains the closed disk D r of radius r and center. Denote α α 2... α n the zeroes of f in the interior of D r repeated according to multiplicity, and suppose that f(). hen log f() = n j= log α j r + log f r dm. (.3)

23 .3. HE ZEROES OF FUNCIONS IN H P 3 Proof. If f is an analytic function, then log f is harmonic except at the zeroes of f. 3 If f is zero free in z ρ and analytic, then log f() = log f ρ dm, (.4) which is the mean-value property A.23 applied on the harmonic function log f. Order the zeros {α j } N n= of f in D r () according to their distance from origo i.e. such that α α n < r and α n+ = = α N = r. Define the function g(z) = f(z) n j= r 2 α j z r(α j z) N j=n+ α j α j z. (.5) Inserting z = into equation (.5) and taking the logarithm gives log g() = log( f() n ) r αj = log f() n + log r α j. (.6) j= On the other hand g has no zeroes in D and hence log g is harmonic, and so log g() = log g r dm, (.7) again by the mean value property. Combining equations (.6) and (.7) gives log n gr dm = log f() log α j r. (.8) Let z = r, then the factors in (.5) for j [n +, N] have absolute value. Since α j = re iθj and z = re iθ it follows that for every n < j N, j= j= α j α j z = =. (.9) z/α j ei(θ θj) Using this and that the first product in equation (.5) equates to one for z = re iθ, we obtain the following expression for log g(re iθ ), log g(re iθ ) = log f(re iθ ) Integrating this expression over gives log fr dm = log N gr dm j=n+ N log e i(θ θj). (.2) j=n+ 2π 2π log e i(θ θ n) dθ. he last integral is evidently independent of θ j and thus zero by Lemma A.. Combining this with equation (.8) completes the proof. 3 Recall that if D is a simply connected domain in C and h a non-vanishing holomorphic function on D then h = e g for some holomorphic function g. So, if D was simply connected we would know that f = e g for some holomorphic g, and then log f = log e g = log(exp(re(g)) = Re(g) and since g is harmonic (g was holomorphic) we are done.

24 4. HARDY SPACES ON HE DISC he next lemma proves a necessary condition on the zeros of a function f in order that f H p for some p. We will later use it to prove that any function in H p may be written as the product of a Blaschke product and a non-vanishing element of H p. Lemma.4 (G. Szegö). Let f H p ( p ) be an analytic function in D such that f and f(). Further, let {α n } n be the zeros of f, listed according to their multiplicities. hen these zeros satisfy the Blaschke condition ( αn ) <. (.2) n Proof. If f has a finite number of zeroes, then the sum is finite and the result follows. herefore, we assume that f has an infinite number of zeroes, since f they converge toward some points in the unit circle. Which is to say lim r z n =. Denote the number of zeroes of f in the closed disk D r by N(r), where r <. Fix K N, and choose r < such that N(r) > K. By Jensen s fomula.3, for each r (, ), we have f() K n= r α n f() N(r) n= ( r α n = exp log ) f r dm <, where the right hand side is bounded as f H p H. Hence, there exists some constant C < such that K n= α n r K f() /C. As the sum now is finite we can let r. Since the inequality holds for all K, we can let K. f() αn C >. n= Using x e x now gives < α n = ( α n ) n= n= exp [ ( α n )] ( ( exp αn )). n= Since e x as x, the inequality above proves that n=( αn ) < as exp( n=( αn ) >. So the Blashke condition (.2) is a necessary condition for the zeroes of an analytic function to belong to a Hardy space H p. Surprisingly enough (.2) is also sufficient condition for the existence of a function f H p, which has zeros only at {α n } n=. Definition. A Blaschke product B(z) is a product of Möbius transformations of the form B(z) := z k α n α n z α n α n z, n We define B(z) = z k, when α = {α j } j is empty. n=

25 .3. HE ZEROES OF FUNCIONS IN H P 5 Proposition.5 (Blaschke product). Let {α n } n= be sequence of complex numbers such that < α α 2 <, α n D, for all n N, satisfying the Blaschke condition (.2). hen the Blaschke product B(z) has only zeroes only at the points α n and a zero of order k at. In addition, B(z) converges uniformly in each disk z R <, we have B(e iθ ) = almost everywhere and B(z) < for all z D. Proof. he function B(z) is the product of the factors b n (z) := a n a n a n z a n z. (.22) Each factor b n has a zero at z = α n inside D, and a pole at z = α outside the closed unit disk D. hus, each factor b n is analytic in D with precisely one zero at α n. Assume that z R then, bn (z) = a n a n z ( a n a n z = an + a n z )( a n ) ( a n an z ) + z a n /a n z a n /a n a n + ( an ). R Since n ( a n ) < it follows that B(z) = n b n(z) converges uniformly in each disk z R <. hat B(z) < is clear since B(z) = a n a n z a n a n z a n a n z a n a n z <, n n as each partial product is less than for z <. Hence, B(e iθ ) by the maximum modulus principle, and the radial limit B(e iθ ) exists almost everywhere (.4). Let f H H, from heorem.2 f r L is increasing and we have the bound f r L f L. (.23) We can apply the inequality above on the function f = B/B n where B n = n k= b k. Since B n (e iθ ) we get (B) r /(B n ) r L B L, (.24) where the slightly convoluted notation (B n ) r = B n (re iθ ) was introduced. As B n (z) B(z) uniformly on z = r we have the inequality B L. (.25) Since B(e iθ ) almost everywhere, this proves that B(e iθ ) = almost everywhere.

26 6. HARDY SPACES ON HE DISC.3.. he Riesz factorization theorem Lemma.4 shows that the zeroes of any nonzero function in in an Hardy space forms a Blaschke product. hus, we can try to divide out the zeros of f by dividing f by the corresponding Blaschke product B. Of course, the resulting quotient g = f/b is again an analytic function in D, and since B has absolute value almost everywhere on the unit circle, we may expect that g have the same H p -norm as the original f. hat this reasoning is indeed correct was proven by F. Riesz in (923) [4]. heorem.6 (F. Riesz). Let f H p, ( p ), f, and let B denote the Blaschke product formed with the zeroes of f in D. If g := f/b, then g H p, g is free of zeroes in D, and g H p = f H p. Proof. From Lemma.4 it is clear that f and B has excactly the same zeroes. Clearly g is then analytic and free of zeroes on D. Let {α n } n be the sequence if zeroes of f in D, and let b n (z) denote the factor of the Blaschke product corresponding to the zero α n as defined in equation (.22). Further, let N B N (z) = b n (z), z D, n= be the partial Blaschke product formed with the first N zeroes of f, and define g N := f/b N. Proposition.5 shows that for every fixed N, we have (B N ) r = B N (re iθ ) uniformly as r. It follows that (g N ) r f and consequently that g N H p = f H p. Since b n (z) < for all n and z D, we have that g (z) g 2 (z) and g n (z) g(z), for every z D. Fixing < r < and applying Lebesgue monotone convergence theorem, one gets lim (g N ) r p H = lim (g N p N ) r p dm = g r p dm = g r p L. N p Since g N is analytic in D and because f r L p f L p (see.2), the left-hand side is bounded from above by f p H for every < r <. Letting r we p obtain g H p f H p. Moreover, since B(z) for all z D, we also have that g(z) f(z) for all z D, this proves that we have equality, i.e that g H p = f H p. Corollary.7. Suppose p <, f H p and again let B be the Blaschke product formed by the zeroes of f. hen there exists a zero-free function g H 2 such that f = B g p/2, (.26)

27 .3. HE ZEROES OF FUNCIONS IN H P 7 and In particular, every f H is a product in which both factors are in H 2 and f p H p = g 2 H 2. (.27) f = gh, (.28) f H = g H2 h H2. (.29) Proof. By heorem.6 f/b H p and f/b H p = f H p a.e. Since f/b has no zeroes in D there exists an analytic ψ D so that e ψ = f/b. Let g = e pψ/2, then g 2 = f/b p, (.3) and so it follows that g H 2 thus, proving equation (.26). Equation (.27) follows directly from integrating equation (.3) over and taking the supremum over r. o prove equation (.28) we can write (.26) in the form f = Bg = f f 2 with f = Bg /2 and f 2 = g /2. Since f, f 2 H 2, we have f H 2 = f 2 H2 = g H2 = f /2 H. Using the last equation twice proves (.29), and we are done Applications of the Riesz factorization theorem Proposition.8 (Mean convergence property). If f H p, ( p < ) then and lim f r L p = f L p, (.3) r lim f r f r L p =. (.32) Proof. We have g 2 L = f p 2 Lp from Corollary.7 so it is enough to prove equation (.3) for H 2. However, this was shown in heorem., and that we may replace the supremum by a limit follows from heorem.2 as the norm is increasing as a function of r. If f(z) = n a nz n, then a n 2 converges when f H 2. From Fatou s lemma A.26, f r f L 2 lim inf f r f ρ = a n 2( r n) 2. (.33) ρ n= Letting r shows equation (.32) for p = 2, since letting r is no problem as the radial limit lim r f(re iθ ) exists for almost every θ,.4. We have proved equation (.3) for all p <, and equation (.32) for p = 2. o deduce (.32) from (.3) we need the following lemma from measure theory.

28 8. HARDY SPACES ON HE DISC Lemma.9 (Duren [5, p. 2]). Let Ω R be a measurable subset, and let ϕ n L p (Ω), p <, and n N. As n suppose that ϕ n (x) ϕ(x) for almost every x Ω and ϕ n (x) p dx ϕ(x) p <. Ω Ω hen, ϕ n (x) ϕ(x) p dx. Ω See Duren [5, p. 2] for proof. proposition.8 now follows from this lemma as f(re iθ ) f(e iθ ) almost everywhere from Fatou s heorem.4 and we have already shown that lim r f r L p = f L p. Lemma.2. Let p and r <. hen, f() p fr p Hp. (.34) Proof. From the mean value theorem A.23 we have f() = f r dm. (.35) Applying the triangle-inequality yields f() f r dm. Using Hölders inequality A.7 with /p + /q = the equation above can be written. ( ) /p ( /q f() f r p dm dθ) q. (.36) Raising both sides of the inequality to the power p completes the proof. With the help of the mean convergence property we are now ready to generalize some properties from heorem. to H p ( p ). Lemma.2 (Point-estimate). Suppose p < and f H p, then f(z) f H p ( z 2 ) /p for all z D. Proof. Following the lines of [49] we consider F r (w) = f ( r z w ) ( z 2) /p zw ( ) 2/p = f ( rϕ z (w)) [ ϕ z(w) ] /p, (.37) zw for < r <. he idea is now to integrate F r (e iθ ) p = f ( rϕ z (e iθ ) ) p ϕ z(e iθ ) with the substitution ϕ z (e iθ ) e iθ such that dθ ϕ z(e iθ ) dθ. So F r (e iθ ) p dθ = f(re iθ ) p dθ = f r p L p f p Hp. (.38)

29 .4. BOUNDARY FUNCIONS 9 From Lemma.2 we get the following inequality for the integral F r (e iθ ) p dθ F r () p = f(rϕ z ()) p ϕ z() = f(z) p ( z 2 ). (.39) Comparing equation (.38) and (.39) completes the proof..4. Boundary functions From heorem.4 we have seen that every function f H p has a nontangential limit f(e iθ ) at almost every boundary point. Let H p () denote the set of boundary functions f(e iθ ). We know from Cauchy s integral formula that a holomorphic function is uniquely determined by its boundary value Proposition A.23, so a Hardy space can be identified with a subspace of the L p (). For the study of Dirichlet series it will be of interest to characterize H p in terms of these boundary functions. Let p from Weierstrass approximation theorem the set of trigonometric polynomials are dense in L p (). hus, a function f in L p () may be written as f(e iθ ) = k Z c k e ikθ, (.4) where c k are the Fourier coefficients. Similarly, H p contains functions on the form n P (e iθ ) = a k e ikθ, (.4) k= where a k are complex constants and these functions will be called polynomials in. he main result is that the polynomials (.4) are dense in H p (D). heorem.22. For every p <, H p () is the closure of the set of polynomials in e iθ. Proof. We begin by considering the analytic function f(z) = n a nz n, f H p (D) and let S N f(z) = N n= a nz n denote the n th partial sum of the aylor series of f at the origin. Proving heorem.22 is the same as proving that for every ε >, there exists a k N, such that S N f f H p ε, for every N k. he idea is to go a small distance λ into the disk, and prove that the result holds for every < λ <. In other words S Nλ f f H p ε. As before we write, f λ (z) = f(λz). Since f H p has a bounded norm on the boundary it follows from Proposition.8 that we can choose an ε such that f λ f H p < ε 2. Similarly, since S N f f(z) uniformly on the circle z = λ, we have by the aylor approximation that for every f(λz), < λ < then S Nλ f f λ p ε 2,

30 2. HARDY SPACES ON HE DISC for sufficiently large enough N λ. Applying Minkowski s inequality A.9 we obtain S Nλ f f H p S Nλ f f λ H p + f λ f H p ε. (.42) hus, proving that the boundary function f(e iθ ) belongs to the L p closure of the polynomials in e iθ. o complete the proof we will show that for if P N (e iθ ) = n k= a ke ikθ is some polynomial such that P N converges to f L p then f H p. he strategy will be to show that all the negative Fourier-coefficients to f is zero. a k = 2π 2π f(e iθ )e ikθ dθ = 2π 2π ( f(e iθ ) P N (e iθ ) ) e ikθ dθ, the last equality follows since the negative Fourier-coefficients of P N are zero. By taking the absolute value and using the triangle inequality we obtain a k 2π f(e iθ ) P N (e iθ ) dθ = f P n L f P n L p. 2π Since P N f the norm above can be made arbitrary small, thus proving that a k = for all k N. Hence, f H p and we are done. o show that heorem.22 does not extend to p = consider the function ( ) z g(z) = exp, z D. z + It is clear that g(z) is analytic for z D because it is the composition of analytic functions, and the only singular point z = lies on the boundary. he function in the exponent (z )/(z +) is a Möbius transformation and maps D onto the left half plane C = {z C: Re(z) < }. Since the exponential function is bounded on C : e σ+it = e σ, this shows that g(z) is also bounded. In fact, on the boundary we have g(e iθ ) =, for almost every θ. Hence, g is Lebesgue integrable and g L. Does this imply that g H? Hardly. Look at g(r) as r. hus, there exists a function in L which does not extend analytically into H (D). Corollary.23. If p, H p is a Banach space. Consequently, H p could be defined as the subspace of those L p functions which all negative Fourier coefficients are equal to zero: Definition. he Hardy space H p for p is the subspace of L p () consisting of functions f such that ˆf(n) = 2π 2π f(t)e int dt = for all n <. H p = { f L p : ˆf(n) = n < }. (.43) Since the polynomials are dense in H p for p <, we will henceforth make no distinction between the spaces H p (D) and H p (). hus, formally defining H p as the closure of all polynomials with respect to the norm on the boundary ( ) f H p = f p p dm. We take the expression above as a radial limit when necessary, that is when p =.

31 .5. CARLEMAN S INEQUALIY 2.5. Carleman s inequality he purpose of this section is to prove Carlemans s inequality, and we offer some historical context for the inequality. he last part is devoted to viewing some generalizations of this inequality. he circle is uniquely characterized by the property that among all simple closed plane curves of given length L, the circle of circumference L encloses maximum area. his property is most succinctly expressed in the isoperimetric inequality A L 2 /4π. (.44) Here A is the area enclosed by a curve C of length L, and where equality holds if and only if C is a circle. here are many known proofs of this fact. More than one idea can be found in the expository paper by Osserman [37], along with a brief histor of the problem. It was Carleman [2] who in 92 gave the first proof based on complex analysis, in the special case of a Jordan domain bounded by a smooth curve. In this section we will see that the theory of the Hardy spaces gives an elementary proof of the inequality. In modern notation equation (.44) may be rewritten as D τ 2 dσ ( 2π 2π τ(e iθ ) dθ) 2 (.45) where τ is a conformal mapping of D onto A, and is known as Carleman s inequality. Here dσ denotes the Lebesgue measure on D normalized so that the measure of D is. In terms of real (rectangular and polar) coordinates, we have dσ = π dx dy = π r dr dθ, z = x + iy = reiθ. Note that in light of heorem.22 the right-handside of equation (.44) is nothing more than τ 2 H. Similarly, we define ( f A p := f p dσ D ) p. (.46) In the passing we mention that the space that contains all analytic functions such that f A p < is called the Bergman space A p, and it has a theory nearly as rich as the Hardy spaces, see Duren [5, p. 25] for a brief overview. We will only make use of the Bergman spaces for its convenient notation though. hus, Carleman s inequality (.45) may be restated as f A 2 f H. (.47) Lemma.24. Let f(z) = n a nz n be analytic in A 2, then ( a j 2 ) 2 f A 2 =. (.48) + j j

32 22. HARDY SPACES ON HE DISC Proof. Recall from Lemma. that zj z k dm = δ jk by the orthogonality of the trigonometric system. hus, z j z k dσ = 2 r i+j z j z k dm dr = 2δ jk 2 + j + k = δ jk + j. D Applying this to equation (.46) the norm of the A 2 space becomes ( ) ( f A 2 = f 2 2 ) ( dσ = a j a k z j z k 2 dσ = D D j,k k a j 2 + j ) 2. By using the Riesz factorization theorem among other results Vukotić presented in [48] a modern and natural way of generalizing (.47). Proposition.25. For p <, every f H p belongs to A 2p, and f A 2p f H p. Proof. We begin by considering the case p = 2 first. Since f is analytic, we can write f(z) = n a n z n, which converges for z D. Squaring this we obtain f 2 = n A n z n where A n = a k a n k. (.49) n Using Lemma.24 on f 2, now gives f 4 A 4 = f 2 2 A 2 = n A n 2 + n = n k= + n a k a n k 2. he last equation can be turned into an inequality by applying the Cauchy-Schwarz inequality n + times f 4 A n ( ) 2 a 4 k a n k 2 = a n 2 = f 4 H. 2 n= k= his proves the case p = 2. Assume that p, if f we are done. If f then f has a finite number of zeroes, and in particular from Riesz factorization theorem.6 we can write f(z) = g(z)b(z), where B is a Blaschke product and g H 2 is zero free in D. Furthermore, from Corollary.7 it is known that g p/2 is in H, since g does not vanish in D, and f H p = g H p. hus, n k f A 2p g A2 p = g p/2 p/2 A 4 g p/2 p/2 H 2 = g H p = f H p, where B(z) < in D from.5 was used to prove the first inequality and the second inequality follows from p = 2. Which completes the proof. he classical isoperimetric inequality now follows directly.

33 .5. CARLEMAN S INEQUALIY 23 Corollary.26. Let G be a Jordan domain with rectifiable boundary of length L( G), and area A(G) hen there holds the inequality A(G) L( G) 2 /4π. Proof. Appealing to the Riemann mapping theorem, we can choose a conformal mapping F of D onto Ω. hen, L( G) = lim L(F ({ z = r})) = lim r r 2π F (re iθ ) dθ = 2π F H. From Duren [5, p. 44] we have that if f(z) maps z < conformally onto the interior of a Jordan curve C, then C is rectifiable if and only if τ H. Also A(Ω) = τ 2 A, so the isoperimetric inequality (.44) follows from f 2 2 A 2 f 2 H applied to f = τ. Generalizations for the weighted Bergman space. While not needed for this thesis, we offer a small digression as to how Carleman s inequality extends to the weighted Bergman space. See the monograph Hedenmalm, Korenblum, and Zhu [23] for further references on the Bergman spaces. Let α > and p <, and define the (weighted) Bergman space A p α(d) as the space of analytic functions in D that are finite with respect to the norm ( ) /p f A p α := f(w) p (α )( w 2 ) α 2 dσ(w) D Here dσ denotes the Lebesgue area measure, normalized so that dσ(d) =. It will be convenient to let dσ α (w) = (α )( w ) α 2 dσ and to let dσ = dm denote the normalized Lebesgue measure on the torus. he following inequality is due to Burbea [, Cor. 3.4] who generalized Carleman s inequality. Proposition.27 (Burbea). Suppose that f H 2, then for every integer k 2 ( ) 2k f A 2k = f(z) 2k dσ k (z) f k H 2. (.5) D Let C α (j) denote the coefficients of the binomial series ( z) α = ( ) j + α C α (j)z j, C α (j) =. (.5) j j= Notice that C (j) = for every j. Identifying C α (j) as the coefficients of the binomial series ( z) α, we find that C αk (j) = C α (j )C α (j 2 ) C α (j k ). (.52) j +j 2+ +j k =j In particular if α is an integer, then C α (j) denotes the number of ways to write j as a sum of α non-negative integers. Hence, C α (j)c β (k) = C α+β (l). j+k=l

34 24. HARDY SPACES ON HE DISC o prove equation (.5) we will need to compute the norm of the weighted Bergman space Lemma.28. Let f = n a nz k be in A 2 α, then f 2 A = a n 2 2 α C α (n). n Proof. Since f A 2 α, we may interchange the integral and summation as needed. f 2 A = f(z) 2 dσ 2 α α (z) D = a n a m r n+m e i(n m)θ (α )( r 2 ) α 2 dσ(w) D n,m = 2(α ) a n 2 r 2n ( r 2 ) α 2 r dr n = a n 2 (α ) n t n ( t) α 2 dt Where the substitution r 2 t was used. Using Corollary A.2 the integral becomes t n ( t) α 2 dt = B(n +, α ) = ( ) n + α = α n (α )C α (n), and we are done. Proof of Proposition.27. Again let f(z) = j a jz j. he idea is to use f 2k = f k 2 and use Lemma.28. f 2k A = 2 a 2k j a jk k C k (j) j j +j 2+ +j k =j ( ) ( ) 2k a j 2 a jk 2 = a j 2 = f 2k H, 2 j j +j 2+ +j k =j j where equation (.52) was used in the second to last inequality. Even though the above proof was relative easy, it is not known whether Proposition.27 holds for non-integer k.

35 .6. HARDY SPACES ON HE POLYDISC Hardy spaces on the polydisc In this section we give a brief introduction to the Hardy spaces of the countably infinite polydisk, H p (D ), which in recent years have recieved considerable interest and study, emerging from the fundamental papers [24, 4]. Much of the renewed interest is due to a simple observation of Bohr [7], which facilitates a link between Dirichlet series and function theory in polydiscs. he standard reference for the Hardy space on the polydisc is the classical monograph [44] by Rudin. We will frequently use polynomials in several complex variables, and for bookkeeping the following multi-index notation is introduced. Definition. An m th order multi-index on C n is the following vector α = (α, α 2,..., α) where α i {,,..., m}. Furthermore α = α +α 2 + +α n = m. For z C n we take z α := z α zα2 2 zαn n. Any m th degree polynomial on n can thus be represented as P (z) = a α z α, α m where we assume that there exists some a α and α = m. Similarly we denote an analytic function on n as with α = (α, α 2, ). f(z) = α a α z α, Definition. Let U be an open subset of C n. A function F : U C is called analytic if it is continuous and analytic in each variable. In one dimension we have studied the unit disk and the unit torus: D = {z C: z < }, = {z C: z = }, where it was clear that D =, and we made no distinction between the spaces. It is natural to consider: D n := {z = (z, z 2,..., z n ) C n : z i D}, n := {z = (z, z 2,..., z n ) C n : z i }. However, if we let z = (,,..., ) then z is on D n but not on, so for n, we see that n D n. hus, some extra care is needed to define H p (D ), since functions in H p (D ) will generally not be well defined in the whole set D. However, similar to the one-dimensional case the radial boundary limit f (z) = lim r f(rz), exists for almost every z d, and we can write f p H p (D d ) = d f p dm d. (.53)

36 26. HARDY SPACES ON HE DISC his means that H p (D d ) is a subspace of L p ( d, m d ). Moreover, again as in the one-dimensional case, for every f H p (D d ) we have lim f f r H r p (D d ) =. (.54) Which implies that the polynomials are dense in H p (D d ). hus, it will be convenient to define the space H p (D d ) as the Banach space completion of the polynomials F (z) = N n= a nz κ(n) in the norm ( ) f H p ( d ) := f p 2 dm d. d As before we make no distinction between H p ( d ) and H p (D d ). A convenient method to obtain equations (.53) and (.54) is to apply the L p -boundedness of the radial maximal function on H p (D d ) for all p >. By Fubini s theorem, the boundedness of the maximal function then reduces to the classical one-dimensional estimate, see [42] for details. o define D, it will be convenient to introduce the set D fin which consists of elements z = {z j } j D such that z j only for finitely many j. It is clear that the function f can be written as a convergent aylor series f(z) = a α z α, z D fin, α N fin and the coefficients c k determine f uniquely. Definition. Let p. he space H p (D ) is the space of analytic functions on D fin obtained by taking the closure of all polynomials in the norm ( ) f H p ( ) := f p p dm. Here dm denotes the Haar measure, we refer to [24] for the details, mentioning only that the Haar measure of is simply the product of the normalized Lebesgue measures in each variable. Lemma.29. For any multi-indices α and β on C d we have d z α z β dm d (z) = δ αβ. (.55) Proof. Recall from Lemma. that in the one dimensional case z k z j dm(z) = δ ij holds for all non-negative integers k and j, by the orthogonality of the trigonometric system. Applying this for each of the d variables completes the proof.

37 .7. HELSON S INEQUALIY he Cole-Gamelin estimate We now wish to show that point evaluations at a point z in D l 2 = {z D : z j 2 < + }, extends continuously to H p ( ). his was first shown by Cole and Gamelin in [4]. Proposition.3 (Cole-Gamelin). Let f H p ( ), where p then ( ) p f(z) z j 2 f Hp (D ), and the inequality is sharp. j j Proof. Let P (z) be a polynomial with z D d, P (z, z 2,, z d ) p. By applying the standard point-estimate.2 to z we obtain P (z, z 2,, z d ) p z 2 P (w, z 2,, z d ) p dm(w ). (.56) Applying.2 to z 2 in equation (.56) gives 2 P (z, z 2,, z d ) p z j 2 P (w, w 2,, z d ) p dm 2 (w, w 2 ), 2 j= by repeating this process and applying the point estimate to each variable we obtain d P (z, z 2,, z d ) q z m 2 P q dm d. d m= Letting d completes the proof. hat this inequality is sharp follows since the point-estimate in one variable is sharp..7. Helson s inequality he purpose of this section will be to generalize Carleman s inequality (.47) ( a k 2 ) 2 f dm, (.57) + k k to. In other words we wish to prove that heorem.3 (Helson s inequality). Given f H (D ) then ( a α 2 ) 2 f H ( + α )( + α 2 ) (D ), (.58) α where α means the unbounded multi-index α = (α, α 2, ).